First list all the factors of 12;
1, 2, 3, 4, 6, 12
now let's see which ones are multiples of 3;
3, 6, 12
so all we have left are numbers 1, 2 and 4.
hope that helps, God bless!
Answer:
im sorry i dont know but this is for a challenge sorry
Step-by-step explanation:
In the ordered pair (2, 12), x = 2 and y = 12.
Substitute these values into each inequality to see if the resulting inequality is true.
y > 2x + 4
12 > 2(2) + 4 Twelve is greater than 2 times 2 plus 4 ...hmmm
12 > 4 + 4
12 > 8 Twelve is greater than 8 ....this is TRUE...go to the next inequality
y < 3x + 7
12 < 3(2) + 7 Twelve is less than 3 times 2 plus 7 ...interesting
12 < 6 + 7
12 < 13 Twelve is less than 13 ... this is TRUE also
Therefore; the answer is b. Yes
It must create a true inequality for BOTH inequalities in order for the ordered pair to be a solution
Answer:
a = 6, b = 3
Step-by-step explanation:
5a - 4 = 2a + 14
3a = 18
a = 6
6 = 2b
b = 3
The answer is C.
x^2 + z = 103x^2 + 22 = 103*Subtract both sides by 22*x^2 + 22 - 22 = 103 - 22x^2 = 81*square root both sides to find x*√x^2 = <span>√81</span>
<span>x = +- 9</span>
